#### To determine

**To evaluate:**

The integral ∫01/3t2-1t4-1 dt

#### Answer

π6

#### Explanation

**1) Concept:**

i) Indefinite Integral

∫1x2+a2dx=1atan-1xa+c

ii) The fundamental theorem of calculus part 2

If f is continuous on a,b, then

∫abfxdx=Fb-F(a)

where F is any antiderivative of f, that is, the function F such that F'=f.

**2) Given:**

∫01/3t2-1t4-1 dt

**3) Calculation:**

∫013t2-1t4-1 dt

Factoring t4-1=(t2-1)(t2+1)

=∫01/3t2-1(t2-1)(t2+1) dt

Cancelling out the common factor,

=∫01/31(t2+1) dt

By using the given indefinite Integral,

∫01/31(t2+1) dt=tan-1t01/3

By applying the fundamental theorem of calculus,

=tan-113-tan-10

=π6-0

=π6

**Conclusion:**

∫01/3t2-1t4-1 dt=π6